Question

Assume that I= E/(R + r),prove that 1/I =R/E + r/E

Answer 1

Answer:

have a graph of 1/current against resistance, which is a straight line of positive gradient. I know that the gradient represents 1/V but I can't work out how the equation 1/I=r/E+R/E relates to y=mx+c. Could somebody please tell me what each part of y=mx+c represents in relation to 1/I=r/E+R/E?

Rearranged equation for Emf= IR + Ir.

Answer 2

Answer:

$\frac{1}{I} = \frac{R}{E} + \frac{r}{E}$ proved.

Explanation:

Given in the  question, I =  $\frac{E}{(R + r)}$.

And from the question we need to prove$\frac{1}{I} = \frac{R}{E} + \frac{r}{E}$.

Therefore, I = $\frac{E}{(R + r)}$ can we written as,

$\frac{1}{I} = \frac{(R +r)}{E}$

• Reciprocal - The definition of reciprocal in mathematics is the inverse of a value or a number.

• If n is a real number, then $\frac{1}{n}$ will be n's reciprocal. Therefore, we must change the number to its upside-down form.

So, $\frac{1}{I} = \frac{(R +r)}{E}$ is the reciprocal of $I = \frac{E}{R +r}$.

Now, $\frac{1}{I} = \frac{(R +r)}{E}$ can be written as, $\frac{1}{I}$ = $\frac{R}{E} + \frac{r}{E}$.

Here, we can see that $\frac{1}{I} = \frac{R}{E} + \frac{r}{E}$

Hence, we proved that $\frac{1}{I} = \frac{R}{E} + \frac{r}{E}$.

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